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Solve the given differential equation.$$\frac{d y}{d x}=\frac{x \sqrt{x^{2}+y^{2}}+y^{2}}{x y}, \quad x>0$$

$y(x)=\sqrt{x^{2} \cdot\left((\ln C x)^{2}-1\right)}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 8

Change of Variables

Differential Equations

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So for this problem, we have a first order differential equation, which I have free written on the top left of the screen that we would like to rewrite this equation in terms of homogeneous variables. Andi, Andi to vote in tomorrow Explicit form. We want to get everything in terms of why we will start this by looking at the school average. So if we factor out X squared underneath the square root we get, the square is equal to the square root of X squared, multiplied by the square root of one plus y squared over X squared. And then this is equal to X times the square root of one plus y squared over X. Glad now we take the positive square root of acts because by the question, X is always positive. So the right hand side of the differential equation now reads X squared. How does he square root of one plus y squared is X squared plus y squared all over X more and we can do one final rewriting by factoring out X squared in both the numerator and a denominator which gives us ey D X is equal to the square root of one plus y squared of the X squared plus y squared over X squared all over. Why over X this we see that we now have the differential equation in the form of being only dependent on why over acts. So it is in fact, homogeneous. We commend. Make the variable transformation off the equals y over X, or why equals v X equivalently. Using the second formulation, we can see that the want the X is equal to V plus X e v the X using the product role. If we substitute this into the left hand side of our equation onset of right hand side, we just substitute in the we get the differential equation being the plus X DVD X is equal to the square root of one plus V squared plus B squared over the we can make an additional simplification by subtracting V from both sides. But on the right hand side notice that if we was to split the fraction into its component forms, the second component V squared over V is simply the itself on V minus three vanishes. So the differential equation is simply x e v. The X is equal to the square root one plus v squared over the This is a separable differential equation. So we will separate the variables. We're in V to the left hand side. Get that. The over the square root of one plus V squared BV is equal to one over x, the X and then to Seoul. So the we need to integrate both science. We will first note that the integral of one over X is simply the natural lager log of X, so that is quite simple. And to integrate the V variables, we will make the substitution off you equals one plus V squared. By differentiating this, we see that to be TV is equal to d you and baffle the integral transforms from the of of a square root of one plus V squared devi into the integral of one over to times square root of you. You. This is simple integration on how solution of the square root of you plus c and then we just we just substitute back into all the variables. Then this is equal to the square root of one plus b squared, plus the integration constant c. So we solved v integral on me nervous solution of the X integral. So the overall equation becomes the square root of one plus V squared. Plus the constancy is equal giving natural log of X. Now, by bringing the sea over to the right hand side and then square in both sides, we get the one plus V squared is equal to the square that the natural log of X plus k, where K is the constant equal to minus c and that's all squared. So we can now simply bring the for want of a right hand side and get our final solution, which is V squared, is equal to the natural log of X plus K all squared minus one. So we now have our solution for they what? We've been asked to find a solution for my not be so We need to substitute variable the equals y of X back into this equation. They're in this. We simply get the Y squared over. X squared is equal. Teoh natural log of X plus K all squared minus one and then multiple it multiplying through by X squared Received that This final solution is all I squared equals X squared all times the square of the natural, Other vex minus K plus K even minus one. This is the solution. The differential equation.

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